A) \[12\times {{10}^{-6}}J\]
B) \[9\times {{10}^{-4}}J\]
C) \[4.5\times {{10}^{-6}}J\]
D) \[2.25\times {{10}^{-6}}J\]
Correct Answer: B
Solution :
Capacitance of capacitor \[=6\mu F=6\times {{10}^{-6}}F\] Initial potential \[{{v}_{1}}=10\,\,V\] Final potential \[{{v}_{2}}=20\,\,V\] The increases in the energy is given by \[\Delta U=\frac{1}{2}C(v_{2}^{2}-v_{1}^{2})\] \[\Delta U=\frac{1}{2}\times 6\times {{10}^{-6}}[{{(20)}^{2}}-{{(10)}^{2}}]\] \[\Delta U=3\times {{10}^{-6}}\times 300\] \[\Delta U=9\times {{10}^{-4}}J\]
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